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I am reading the datasheet of the LM3409, specifically section 9 "Application and implementation". My intent is to understand how to calculate the values of the components needed for my specific application.

I aim to drive a single LED with a maximum constant current of 2.7A, from a 42V PSU, and use the analog dimming feature of the LM3409 to control the brightness (similar to the circuit in section 9.2.2).

Most of what is explained in the datasheet is sufficiently clear to me, except for how to determine the inductor ripple current (ΔiL-PP) and the LED ripple current (ΔiLED-PP).

Chapters 8.3.4 (Inductor Current Ripple) and 9.1.3 (LED Ripple Current) seem to imply that I should be choosing these values, but they give me no clear indication as to how to choose them.

Can anyone help me understand how to choose these values ?

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2 Answers 2

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You could use WebBench to calculate the values for you. Set up the parameters then go into the BOM and select the inductor that best meets your efficiency (DCR), physical size, and cost.

I used a supply of 40-43V and Vf of 35V assuming you are using a CoB similar to a Luminus CHM-27

WebBench Parameters
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The WebBench design below has a IL-PP ripple current of 840 mA (32%) and efficiency of 97%.

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Once you have your nominal switching frequency, you can then choose the inductor ripple current.

Very similar to Section 9.2.2.2.2

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source: AN-1953 LM3409HV Evaluation Board

For additional understanding see section 2.3 Critical Inductance of this TI Application Report: Understanding Buck Power Stages in Switch mode Power Supplies

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  • \$\begingroup\$ As of today WebBench no longer supports LM3409 \$\endgroup\$
    – Travis Su
    Commented Feb 14, 2023 at 8:11
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The Inductor ripple current should be chosen according to your application . The value will depend on the available size of your driver and the range of input voltage and output load.

In your case the input voltage (42V) and output load (LED's ) are fixed.

Choosing a high current ripple results in a high peak current which requires a inductor with a high saturation current. Also, the core losses, which are dependent on the AC part will also be higher.

Choosing a low current ripple requires a bigger inductor. This translates to a higher resistance since you need more turns , thus resulting in higher I^2 * R losses.

So, the optimum value of the inductor needs to be optimized. A good starting point is 30-40%.

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The luminous flux of an LED is mostly linear, but at a certain point some of the current just leads to heat. Thus you need to calculate based on the LED's you plan to in order to not over stress them.

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